Wind Power B/C

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Re: Wind Power B/C

Post by chalker »

soyuppy wrote:
windu34 wrote:
mkfiddler11 wrote:I agree that the voltage output depends on the setup. Our team ends up getting lower voltages at the tournament even though the resistors used are higher, may be the motor used is of lower power than what we use. It would be helpful if someone can post the results from 2016 nationals.
Um okay how well do you know circuits...higher resistance means lower voltage because its harder for the turbine to spin the motor shaft
no...V=I*R. If R increase, assuming I remain the same, then V increase. V is the voltage drop. Or you can test it yourself. Just use higher resistance in you setup on the same blade and motor. See what happen.
There's a BIG assumption you made - I remaining the same. In reality, I won't usually remain the same in this circuit. Windu34 is correct in that by putting a higher resistor in the circuit, it makes it harder for the turbine to spin the motor. The reasoning behind this is due to something called EMF (https://en.wikipedia.org/wiki/Electromotive_force) and Faraday's Law of Induction (https://en.wikipedia.org/wiki/Faraday%2 ... _induction). The best way to think about this is that the blades generate a force on the shaft. That force has to overcome the 'physical resistance' of the motor to turn the shaft. That 'physical resistance' is due to a combination of the magnetic 'flux' of the motor windings and the electrical resistance of the circuit. Increase the electrical resistance, and the same amount of force will have more 'physical resistance' to overcome, resulting in less output (e.g. Power) from the generator (i.e. less current and voltage).

In many electrical engineering classes, the hydraulic analogy (https://en.wikipedia.org/wiki/Hydraulic_analogy) is used to help explain things, by comparing electrical circuits to water plumbing. In this situation, think of the generator as a water pump. By increasing the circuit resistance, you are effective making the hose connected to the pump a smaller diameter. If you hook up a smaller diameter hose to a pump, the amount of water you get out of the pump will go down (e.g. the current decreases), but that won't be necessarily offset by the water coming out of the hose being more forceful (e.g. the voltage), hence the net power is reduced.

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Re: Wind Power B/C

Post by chalker »

soyuppy wrote:Just post similar question to SO inc.
When you buy a bundle of CD/DVD, usually come in 25/50/100, you get the plastic filler. Can this plastic filler be used for blade? Has the same dimension as CD/DVD, but it's much lighter.
As usual, this isn't the place for official statements / clarifications. That said, if you asked 100 random people off the street whether that plastic filler was a CD, what do you think the majority of them would say? I'd propose they'd say No, it isn't a CD.

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Re: Wind Power B/C

Post by LIPX3 »

I've looked over the rules, and I haven't found anything that says you can't use a vertical axis turbine. Has anyone seen someone do this? I think it would be a way to increase surface area of the blades, as you can go as far in front of the CD as you want.
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Re: Wind Power B/C

Post by ashmmohan »

Unome wrote:
soyuppy wrote:
mkfiddler11 wrote:Has any of you been to an invitational and know of top high and low voltage scores at the event? Please post the resistance used in the setup. Thanks!
I just helped ran an invitational event.

CD motor: https://www.amazon.com/gp/product/B00OT ... UTF8&psc=1
Load R: 10 Ohm

Top blade = 700mV+ on high, 500mV+ on low
Low blade = 80mv on high, 70mv on low
Was this B or C? (I'll go ahead and assume this was at Mira Loma yesterday)
Keep in mind that division B can have their blades extend as far as 20 cm from the center of the CD. 700mV is pretty high for any setup, but the 20 cm blades can increase results tremendously.
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Re: Wind Power B/C

Post by chalker »

LIPX3 wrote:I've looked over the rules, and I haven't found anything that says you can't use a vertical axis turbine. Has anyone seen someone do this? I think it would be a way to increase surface area of the blades, as you can go as far in front of the CD as you want.
Read rule 2.d.iii closely, particularly the bolded part (which indicates it's new this year). We explicitly put that in place to prevent VAWT designs. In general, VAWT designs are much harder to make and put much more torque / stress on the motor / CD hub mount.

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Re: Wind Power B/C

Post by LIPX3 »

chalker wrote:
LIPX3 wrote:I've looked over the rules, and I haven't found anything that says you can't use a vertical axis turbine. Has anyone seen someone do this? I think it would be a way to increase surface area of the blades, as you can go as far in front of the CD as you want.
Read rule 2.d.iii closely, particularly the bolded part (which indicates it's new this year). We explicitly put that in place to prevent VAWT designs. In general, VAWT designs are much harder to make and put much more torque / stress on the motor / CD hub mount.
I see that now. I was thinking most setups wouldn't even allow a vertical axis to mount.
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Re: Wind Power B/C

Post by mkfiddler11 »

chalker wrote:
soyuppy wrote:
windu34 wrote: Um okay how well do you know circuits...higher resistance means lower voltage because its harder for the turbine to spin the motor shaft
no...V=I*R. If R increase, assuming I remain the same, then V increase. V is the voltage drop. Or you can test it yourself. Just use higher resistance in you setup on the same blade and motor. See what happen.
There's a BIG assumption you made - I remaining the same. In reality, I won't usually remain the same in this circuit. Windu34 is correct in that by putting a higher resistor in the circuit, it makes it harder for the turbine to spin the motor. The reasoning behind this is due to something called EMF (https://en.wikipedia.org/wiki/Electromotive_force) and Faraday's Law of Induction (https://en.wikipedia.org/wiki/Faraday%2 ... _induction). The best way to think about this is that the blades generate a force on the shaft. That force has to overcome the 'physical resistance' of the motor to turn the shaft. That 'physical resistance' is due to a combination of the magnetic 'flux' of the motor windings and the electrical resistance of the circuit. Increase the electrical resistance, and the same amount of force will have more 'physical resistance' to overcome, resulting in less output (e.g. Power) from the generator (i.e. less current and voltage).

In many electrical engineering classes, the hydraulic analogy (https://en.wikipedia.org/wiki/Hydraulic_analogy) is used to help explain things, by comparing electrical circuits to water plumbing. In this situation, think of the generator as a water pump. By increasing the circuit resistance, you are effective making the hose connected to the pump a smaller diameter. If you hook up a smaller diameter hose to a pump, the amount of water you get out of the pump will go down (e.g. the current decreases), but that won't be necessarily offset by the water coming out of the hose being more forceful (e.g. the voltage), hence the net power is reduced.
Thank you all for the responses. Our team experimented with different resistors to check the voltage results. With blades and the stand setup constant, they measured higher voltages for higher resistance in the circuit (10 ohm vs 5 ohms). The voltage output is almost double for both high power and low power of the fan. Whereas, at a tournament their voltage dropped when a 6.8 ohm resistor was used. I feel that the motor power at the tournament must be lower than our setup.
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Re: Wind Power B/C

Post by chalker »

mkfiddler11 wrote:
chalker wrote:
soyuppy wrote: no...V=I*R. If R increase, assuming I remain the same, then V increase. V is the voltage drop. Or you can test it yourself. Just use higher resistance in you setup on the same blade and motor. See what happen.
There's a BIG assumption you made - I remaining the same. In reality, I won't usually remain the same in this circuit. Windu34 is correct in that by putting a higher resistor in the circuit, it makes it harder for the turbine to spin the motor. The reasoning behind this is due to something called EMF (https://en.wikipedia.org/wiki/Electromotive_force) and Faraday's Law of Induction (https://en.wikipedia.org/wiki/Faraday%2 ... _induction). The best way to think about this is that the blades generate a force on the shaft. That force has to overcome the 'physical resistance' of the motor to turn the shaft. That 'physical resistance' is due to a combination of the magnetic 'flux' of the motor windings and the electrical resistance of the circuit. Increase the electrical resistance, and the same amount of force will have more 'physical resistance' to overcome, resulting in less output (e.g. Power) from the generator (i.e. less current and voltage).

In many electrical engineering classes, the hydraulic analogy (https://en.wikipedia.org/wiki/Hydraulic_analogy) is used to help explain things, by comparing electrical circuits to water plumbing. In this situation, think of the generator as a water pump. By increasing the circuit resistance, you are effective making the hose connected to the pump a smaller diameter. If you hook up a smaller diameter hose to a pump, the amount of water you get out of the pump will go down (e.g. the current decreases), but that won't be necessarily offset by the water coming out of the hose being more forceful (e.g. the voltage), hence the net power is reduced.
Thank you all for the responses. Our team experimented with different resistors to check the voltage results. With blades and the stand setup constant, they measured higher voltages for higher resistance in the circuit (10 ohm vs 5 ohms). The voltage output is almost double for both high power and low power of the fan. Whereas, at a tournament their voltage dropped when a 6.8 ohm resistor was used. I feel that the motor power at the tournament must be lower than our setup.
I stand corrected. This is very counterintuitive, but indeed higher resistance in this configuration will generally result in higher measured voltages (versus what would happen in a full scale generator that takes into account active and reactive power - https://en.wikipedia.org/wiki/AC_power#Reactive_power). I think the reason for this is that I forgot to take into account that these motors have internal resistances that are a few ohms and generate DC power. Hence, we are effectively forming a voltage divider circuit (https://en.wikipedia.org/wiki/Voltage_divider). By increasing the resistance, we are increasing the proportion of the voltage that goes to the external resistor versus the internal resistor, hence the voltage goes up (to a limit).

This is one of the reasons I love SO... there's always an opportunity to learn something new and be surprised when the practical application of something doesn't align with your theoretical understanding. Thanks for bringing this up and pointing out the results you were seeing!

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Re: Wind Power B/C

Post by ashmmohan »

On our setup, we have a 5 ohm and 25 ohm resistor connected to the motor and a switch to change the resistance back and forth. Works pretty well.
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Re: Wind Power B/C

Post by Alex-RCHS »

Have you been noticing certain designs working well at one resistance and poorly at another? Or is it the case that (generally) if a design works well at 5 ohms it also works well at 25 ohms?

Of course, "working well" is relative to other designs.
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